Mathematics Department, UCLA
T. Richthammer
winter 09, sheet 9
Feb 27, 2009
Homework assignments: Math 170A Probability, Sec. 1
109. Repeat problem 104, now assuming that the arrival time
X
of the bus is an exponential
RV with parameter
α
= 1
/
15.
Answer:
(a)
P
(
X >
10) =
R
∞
10
αe

αx
dx
=
e

10
/
15
≈
0
.
51.
(b)
P
(
X
≥
25

X >
15) =
P
(
X >
10)
≈
0
.
51. (memoryless!)
110. Repeat problem 105, now assuming the road has inﬁnite length, and
X
is exponentially
distributed with parameter
α
.
Answer:
E
(

X

a

) =
R
∞
0
α

x

a

e

αx
dx
=
R
a
0
(
a

x
)
αe

αx
dx
+
R
∞
a
(
x

a
)
αe

αx
dx
=
a
+
1
α
(
e

αa

1) +
1
α
e

αa
=
a
+
2
α
e

αa

1
α
=
f
(
a
), using
R
(
x

a
)
αe

αx
dx
=

(
x

a
)
e

αx

1
α
e

αx
(integration by parts). Minimizing
f
(
a
):
f
±
(
a
) = 0 gives
a
=
ln 2
α
, and we get
f
(
a
) =
ln 2
α
.
111.* In the lecture we have argued that under certain assumptions the number of events in
an interval [0
, t
] is a Poisson RV with parameter
αt
by using a discretization argument
and the Poisson approximation of binomial RVs. From this we have concluded that the
waiting time
Y
for the ﬁrst event is an exponential RV with parmeter
α
. We can also
argue directly, using the discretization and a corresponding approximation theorem on the
waiting times rather than on the number of successes: We divide the interval [0
,
∞
) into
pieces of equal length 1
/n
, and let
Y
n
denote the waiting time for the ﬁrst interval that
contains an event. Show the following:
(a) The assumptions from 6.3(d) imply that
Y
n
is a geometric RV with parameter
α/n
.
(b)
Y
n
/n
≈
Y
for large
n
.
(c) Whenever
Y
n
is a geometric RV with parameter
p
n
such that
np
n
→
α >
0 for
n
→ ∞
, the distribution function of
Y
n
/n
converges to the distribution function of
an exponential RV with parameter
α
at every point
c
.
(d)
Y
has to be an exponential RV with parameter
α
.
Answer:
(a) As in 6.3(d) we see that the RVs
X
i
(where
X
i
= 1 if an event occurs in the
i
th
interval, and
X
i
= 0 otherwise) form a Bernoulli sequence with parameter
p
=
α/n
.
Y
n
is
the waiting time for the ﬁrst success in the Bernoulli sequence and thus a geometric RV
with parameter
p
=
α/n
.
(b) If the ﬁrst success occurs in the
i
th interval (which has endpoints
i

1
n
and
i
n
), then
Y
n
=
i
and
Y
is in this interval, so

Y

Y
n
/n
 ≤
1
n
.
(c)
F
Y
n
/n
(
c
) =
P
(
Y
n
≤
cn
) =
P
(
Y
n
≤
c
n
), where
c
n
:=
±
cn
²
. We have
P
(
Y
n
≤
c
n
) =
∑
c
n
k
=1
p
n
(1

p
n
)
k
=
p
n
1

(1

p
n
)
cn
+1
1

(1

p
n
)
= 1

(1

p
n
)
c
n
+1
= 1

[(1

p
n
)
n
]
c
n
/n
, and in the
Poisson approximation theorem we have shown (1

p
n
)
n
→
e

α
, and we have
c
n
/n
=
±
cn
²
/n
→
c
. Thus
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 Winter '10
 RICHTHAMMER
 Normal Distribution, Binomial distribution, yn, normal RV, exponential RV

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